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I want to show:

If $\mathfrak{g}$ is a Lie algebra that has an abelian subalgebra $\mathfrak{h}$ such that $\mathfrak{g}$ has a Cartan decomposition $\mathfrak{g}=\mathfrak{h}\oplus(\bigoplus_{\alpha \in \mathfrak{h}^*} \mathfrak{g}_\alpha)$, then $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is a Cartan subalgebra.

$\mathfrak{g}_\alpha$ denote the root space; for any $H\in \mathfrak{h}$ and $X\in \mathfrak{g}_\alpha$, $\text{ad}(H)(X)=\alpha(H)\cdot X$. If $R$ denotes the set of roots, you can also verify that:

  1. $\dim \mathfrak{g}_\alpha=1$
  2. $R$ generates a lattice $\subset \mathfrak{h}^*$ with rank $=\dim \mathfrak{h}$.
  3. If $\alpha \in R$, then $-\alpha \in R$.

I also know $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]\neq 0$ and $[[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}],\mathfrak{g}_\alpha]\neq 0$ for $\alpha$ being a root.

For the first part I am not sure how to show $\mathfrak{g}$ is semisimple. Could anybody walk me through this? For the second part I need $\mathfrak{h}$ is a maximal toral abelian subalgebra. Clearly it is maximal abelian because $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]\neq 0$ for all the other root spaces. How do I show it is a maximal toral subalgebra?

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What properties does this decomposition satisfy? Any Lie algebra has a direct sum decomposition (as a vector space). If the only property be that this is a root space decomposition then this is also not sufficient to be semisimple since any abelian Lie algebra $\mathfrak g$ has such a decomposition (with $\mathfrak h = \mathfrak g$ and $\mathfrak g_\alpha = 0$ for all $\alpha$) but are not semisimple –  Eric O. Korman Mar 11 '13 at 13:15
    
Clarifications: Is the direct sum ranging over $\alpha \in \mathfrak{h}^*$? Are you assuming $\mathfrak{g}_0 = \mathfrak{h}$? Even more basically, are you assuming $\mathfrak{g}_\alpha = \{ g \in \mathfrak{g} \;|\; [h,g]=\alpha(h)g \}$? As Eric pointed out, your question lacks a lot of needed detail. –  Bill Cook Mar 12 '13 at 1:49
    
I had thought that the notation was standard, but now I know this isn't the case (sorry). The decomposition $\mathfrak{g}=\mathfrak{h}\oplus(\bigoplus \mathfrak{g}_\alpha)$ is a Cartan decomposition, and indeed $\alpha \in \mathfrak{h}^*$. Yes, I am assuming that the root space is denoted $\mathfrak{g}_\alpha$ (with the standard properties: $\dim \mathfrak{g}_\alpha=1$; the set of roots $R$ generates a lattice $\subset \mathfrak{h}^*$ with rank $=\dim \mathfrak{h}$; $\alpha \in R$ implies that $-\alpha\in R$). This is also Exercise 14.34 in Fulton's Representation Theory. –  user64913 Mar 13 '13 at 2:29
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